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Let $\lambda_1, \lambda_2, \lambda_3$ denote the eigenvalues of the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos  t & \sin  t \\ 0 & -\sin  t & \cos  t \end{pmatrix}$.

If $\lambda_1+ \lambda_2+\lambda_3=\sqrt{2} +1$ then the set of possible values of $t, - \pi \leq t < \pi$, is

  1. Empty set
  2. $\{ \frac{\pi}{4} \}$
  3. $\{ - \frac{\pi}{4}, \frac{\pi}{4} \}$
  4. $\{ - \frac{\pi}{3}, \frac{\pi}{3} \}$

 

asked in Linear Algebra by Loyal (4k points)  
edited by | 95 views

2 Answers

+1 vote

Answer C)

 

1-⋋  0        0

0    cos t-⋋ sin t

0     -sin t    cos t - ⋋

 

 

solving (1-⋋) {(cos t - ⋋)2 - sin2⋋}=0................i

⋋=1

⋋1+⋋2+⋋3=1+√2

if ⋋1=1

⋋2+⋋3=√2

 

from equation i we can say (cos t - ⋋)2 - sin2⋋=0

(cos t - ⋋) =(+/-) sin⋋

⋋= cos t -sin t

t=(⊼/2⨉2 + ⊼/4)  as range of t is -⊼<t<⊼

So, ans should be C)

answered by Veteran (53.9k points)  

srestha I think Equation (1) should be

$(1-\lambda ) [{(cos\ t -\lambda )^{2} + sin^{2}} \ t] =0$

yes and answer will be a complex number, rt?

srestha check this.

We know that Sum of eigen values = trace of the matrix

Here $\lambda_{1}+\lambda _{2}+\lambda _{3}= \sqrt{2}+1$

$1+2 \ cos\ t= \sqrt{2}+1$

$2 \ cos\ t= \sqrt{2}$

$cos \ t = \frac{1}{\sqrt{2}}$

$t = cos^{-1}(\frac{1}{\sqrt{2}})$

$t =-\frac{\pi }{4}\ , \ \frac{\pi }{4}$

 

srestha I think this method is more easy

yes, that is correct way :)
+1 vote
We know that Sum of eigen values = trace of the matrix

Here $\lambda_{1}+\lambda _{2}+\lambda _{3}= \sqrt{2}+1$

$1+2 \ cos\ t= \sqrt{2}+1$

$2 \ cos\ t= \sqrt{2}$

$cos \ t = \frac{1}{\sqrt{2}}$

$t = cos^{-1}(\frac{1}{\sqrt{2}})$

$t=\{-\frac{\pi}{4},\frac{\pi}{4}\}$

 

Hence,Option(C)$\{-\frac{\pi}{4},\frac{\pi}{4}\}$.
answered ago by Veteran (31.1k points)  


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